Question

 $sqlinsert="SELECT * FROM imagetable "; 
    $result = mysqli_query($con, $sqlinsert);

    echo "<table border='1'>";
    echo "<tr>
              <th>ID</th>            
              <th>School Code</th>
              <th>Category</th>
              <th> School name </th>
              <th> School Address </th> 
              <th> School name </th>

          </tr>";

    while($row = mysqli_fetch_assoc($result)){
        $id=  $row['ID'];
        echo "<tr>
                  <td>";
                  echo $row['ID'];
                  echo "<td>";
                  echo $row['category'];
                  echo "<td>";
                  echo $row['sname'];
                  echo "<td>";
                  echo   $row['sadd'];
                  echo "<td>";
                  echo   $row['filename'];

                  echo "<td>";
                 ?>             


           <form name="form" method="POST" action="parentssubmit.php">
         <input  type="hidden" value="<?php echo  $id;?>" name="search">
         <input type="submit"  value="View" name="View">
       </form>
    <?php   } 

         echo "</td></tr>";
    echo "</table>";
    ========================================================================
    if (isset($_POST['View']) ){
     $pdf=new fpdf();
    $pdf->ADDPage();
    $pdf->setfont('Arial','B', 16);  
    $pdf->Cell(40,10, 'sname',1,0,'c');
    $pdf->Cell(40,10, 'sadd',1,0,'c');
    $pdf->Cell(40,10, 'category',1,0,'c');
    $pdf->Cell(40,10, 'scode',1,0,'c');


         $con = mysqli_connect("localhost","root","","school");
           if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }


       $file_filename=$_FILES['file']['name'];

       $target_path  =  "Newfolder1";


     $image_path = $target_path . DIRECTORY_SEPARATOR . "filename";

        $image_format = strtolower(pathinfo('filename', PATHINFO_EXTENSION));
        $sql="SELECT * FROM imagetable WHERE ID= '$_POST[search]'";
        $result = mysqli_query($con, $sql);
          if(!mysqli_query($con, $sql)){
                    die( "Could not execute sql: $sql"); 
                     }

    // build the data array from the database records.
    While($row = mysqli_fetch_array($result)) {
           $pdf->Cell(40,10, $row['sname'],1,0,'c');
            $pdf->Cell(40,10, $row['sadd'],1,0,'c');
            $pdf->Cell(40,10, $row['category'],1,0,'c');


    $pdf->Image($image_path,50,100,50,20,$image_format);

           $pdf->ln();}
    }
    $pdf->output();

这是两页，错误出现未定义索引：

文件未定义的索引：文件所以请尝试指出错误，因为当我添加图像字段时出现主要错误然后错误开始弹出。

我正在尝试从数据库中获取图像并尝试以PDF格式显示它。

Answer 1

首先，您应该关闭所有<tr>和<td>代码。其次，您没有向表单发送任何文件，因此您收到此Undefined index: file错误，因此请删除这些行，

$file_filename=$_FILES['file']['name'];
$target_path  =  "Newfolder1";
$image_path = $target_path . DIRECTORY_SEPARATOR . "filename";
$image_format = strtolower(pathinfo('filename', PATHINFO_EXTENSION));

因此，在用户点击view按钮后，您应该处理表单并显示如下图像，

// your code

if (isset($_POST['View']) ){
    $pdf=new fpdf();
    $pdf->ADDPage();
    $pdf->setfont('Arial','B', 16);  
    $pdf->Cell(40,10, 'sname',1,0,'c');
    $pdf->Cell(40,10, 'sadd',1,0,'c');
    $pdf->Cell(40,10, 'category',1,0,'c');
    $pdf->Cell(40,10, 'scode',1,0,'c');


    $con = mysqli_connect("localhost","root","","school");
    if (mysqli_connect_errno()){
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $sql="SELECT * FROM imagetable WHERE ID= '$_POST[search]'";
    $result = mysqli_query($con, $sql);
    if(!mysqli_query($con, $sql)){
        die( "Could not execute sql: $sql"); 
    }

    $row = mysqli_fetch_array($result);
    $image_path = $row['file'] . DIRECTORY_SEPARATOR . $row['filename'];  // path should be like this, process/upload/8/cdv_photo_001.jpg
    $image_format = strtolower(pathinfo($image_path, PATHINFO_EXTENSION));
    $pdf->Cell(40,10, $row['sname'],1,0,'c');
    $pdf->Cell(40,10, $row['sadd'],1,0,'c');
    $pdf->Cell(40,10, $row['category'],1,0,'c');
    $pdf->Image($image_path,50,100,50,20,$image_format);
    $pdf->ln();
    $pdf->output();
}

// your code

fpdf错误和视图表错误

1 个答案: